Fluid Pressure Drop Calculator

Pressure Drop Formula:

\[ \Delta P = f \times \left(\frac{L}{D}\right) \times \left(\frac{\rho \times V^2}{2}\right) \]

Friction Factor (f):

Pipe Length (L):

Pipe Diameter (D):

Fluid Density (ρ):

kg/m³

Fluid Velocity (V):

m/s

Pressure Drop (ΔP):

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1. What is a Fluid Pressure Drop Calculator?

Definition: This calculator estimates the pressure loss in a pipe due to friction between the fluid and pipe walls.

Purpose: It helps engineers and technicians determine the energy loss in piping systems for proper pump sizing and system design.

2. How Does the Calculator Work?

The calculator uses the Darcy-Weisbach equation:

\[ \Delta P = f \times \left(\frac{L}{D}\right) \times \left(\frac{\rho \times V^2}{2}\right) \]

Where:

\( \Delta P \) — Pressure drop (Pascals)
\( f \) — Darcy friction factor (dimensionless)
\( L \) — Pipe length (meters)
\( D \) — Pipe diameter (meters)
\( \rho \) — Fluid density (kg/m³)
\( V \) — Fluid velocity (m/s)

Explanation: The equation calculates the energy loss due to friction between the moving fluid and pipe walls.

3. Importance of Pressure Drop Calculation

Details: Accurate pressure drop calculations ensure proper pump selection, energy efficiency, and system performance in piping networks.

4. Using the Calculator

Tips: Enter the friction factor (default 0.02 for turbulent flow), pipe dimensions, fluid properties, and velocity. All values must be > 0.

5. Frequently Asked Questions (FAQ)

Q1: How do I determine the friction factor?
A: For turbulent flow, use Moody chart or Colebrook equation. For laminar flow (Re < 2000), f = 64/Re.

Q2: What's a typical water density value?
A: Pure water at 20°C has density of 998 kg/m³, often rounded to 1000 kg/m³ for calculations.

Q3: Does this include minor losses?
A: No, this calculates only major (friction) losses. Add minor losses from fittings separately.

Q4: What velocity range is typical?
A: For water, 1-3 m/s is common in pipes to balance pressure drop and erosion concerns.

Q5: Can I use this for gases?
A: Yes, but for compressible gases with significant pressure drop, more complex methods are needed.